I'm working through a proof from lecture notes in an algebra class that for all $n$, the $n$-th cyclotomic polynomial $\Phi_n$ has integer coefficients. The proof proceeds as follows:
- Write $x^n - 1 = \Phi_n \cdot g$ with $g \in \mathbb{Q}[x]$.
- There are $0 \ne a, b\in \mathbb{Z}$ such that $a\cdot\Phi_n, b\cdot g \in \mathbb{Z}[x]$ and are both primitive.
- By Gauss's lemma, $ab(x^n - 1) = (a \Phi_n)(b g)$ is primitive, and so $ab = 1$, and as such $a = b = \pm 1$, and so $\Phi_n \in \mathbb{Z}[x]$
My issue is with step 2. I understand how you can justify this if you relax it to $a,b \in \mathbb{Q}$, but we need $a,b \in \mathbb{Z}$ for the proof. I've been looking for quite a bit, and despite this being a relatively simple lemma, I can't find a proof of it and am starting to wonder if it's even true, especially given that other proofs of the same thing seem to be much longer.
You are right with the issue in step 2. There is a simple counterexample: $$ g(x) = 2x + \frac{2}{3} $$ However, $\Phi_n(x)$ has leading coefficient 1, and so does $g(x)$ since $x^n-1 = \Phi_n(x)\cdot g(x)$. Therefore in this particular case of the proof, $a,b\in\mathbb{Z}$ is correct.